LIBRARY OF CONGRESS. 



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UNITED STATES OF AMI itIC A. 






/ 

GEOMETRICAL 

Square Root; 

A CIRCLE QUADRATURES 



OTHER PROBLEMS. 



SIT ZETODRIMI^lvr OBABB. 



I :> 1879. Jrf. 



CHICAGO: 

187 South Green Street. 
1879. 



Entered according to Act of Congress, in the year 1S79, By 

NORMAN CRABB, 
In the Office of the Librarian of Congress at Washington. 



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PREFACE. 

The author of this little work, thinking 
that the correct method of solving these 
twelve problems might be of some bene- 
fit to the public, has been induced to 
publish them, as he believes that none of 
them have ever been correctly solved 
Every attempt to solve these problems 
which has come to my notice, has proved 
a failure, for the reason that a fact can- 
not be found by the use of an imperfect 
theory. For instance, if the rule of square 
root be used by figures (not adopting my 
theory) to get the root of nineteen and 
one-half, it proves a failure. Hence it is 
an imperfect theory, to sovle this problem. 
The Author, 

Norman Crabb. 



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Problem No. i. 

FIGURE F. 

Fig. F shows the method of square root 
by geometry. 

Let the large square A, F, G, E, rep- 
resent four. Attach the square H, I, l r 
G, of one-fourth the area of large square r 
Attach to that the small square Z, I, M, 
N, which is one-half the area of square 
G, H, I, I. Set your dividers at G, open 
to Z, swing around to O, raise the per- 
pendicular line O, C, in height equal to 
G, O. Then draw the line C, P, that forms 
the square G, P, C, O, which equals the 
two small squares. Consequently it must 
be the square root of one and one-half. 
Then set off the distance from E to D, 
equal to the distance from G to O, draw 
the line D, C, which is the root of all of 
the squares. Form the square D, C, B, 
A, and to prove that it is the root of five 
and one-half, please observe that it leaves 
out two right-angled triangles D, E, A, 
and D, C, O, and takes in two triangles 
P, C, B, and A, B, F, which equals the 
two triangles left out These five squares 



6 

may represent, say each twenty-five. And 
small square twelve and one-half. This 
would make it the square root of one 
hundred and thirty-seven and one-half. By 
this method any number including frac- 
tions may be square rooted by drawing 
to a scale, and the proper arrangements 
of the squares. The area of squares may 
de divided and multiplied into any de- 
sired size or fractions. 

To bring them into shape for square 
rooting, follow the rules for figure E. 



Problem No. 2. 

FIGURE B. 

Fig. E, shows the method of multiply- 
ing and dividing the area of squares or 
equilateral triangles. 

Let A, B, C, D, be a given square. To 
halve the area, draw the diagonal lines A, 
F, D, B. Either of the diagonal lines 
from the corners of square to center F, 
will be the side of a square of half the 
area of large square. Form the square 
A, H, D, F. Half of small square occu- 
pies one-fourth of large square. The other 
half being equal, must together be equal 
to half of large square, or either of the 
diagonals from corner to corner will be 
the side of a square of twice the area of 
large square. The triangle in large square 
being an equilateral triangle, and one side 
being equal to one side of square, it must 
be the largest equilateral triangle that can 
be drawn in that square. Now the small 
triangle being the largest equilateral tri- 
angle that can be drawn in small square, 
it follows that it must occupy the same 
proportion of the area of small square as 



the large triangle does of the large square. 
The small square being half the area of 
large square, the small triangle must be 
half the area of large triangle. 



Problem No. 3. 

FIGURE I. 

Fig. I, is an equilateral triangle halved 
in two different ways, G, H, being one 
halving line and I, O, the other. First we 
will halve it with the line H,G. That can 
be done by taking half the square of the 
altitude from the angle C to N. Process : 
Set your dividers at D, open to C, swing 
round to E, intersecting base line ; from 
the intersection draw the line C, E, then 
draw the line D, and F, at right angles 
with the line C, E, and intersecting the 
angle at D. This line is half of the 
square of the altitude. The length of 
this line taken from angle C, is the halving 
point Draw the line G, H, parallel to the 
base and through the point, and you have 
the triangle halved. Process: For halving 
from a given point, in one side of triangle; 
first draw a rectangle, one side to equal 
the height of triangle, and the other to 
equal half of one side of triangle. This 
forms a rectangle, the area of which is 
equal to the area of triangle to be halved. 
Any triangle of the largest size that can 



IO 

be drawn in this rectangle, no matter 
what shape, will equal the half of said 
triangle to be halved. But we have two 
dimensions; from the point O to the 
angle C is one, the perpendicular line 

0, O, is another; now this rectangle must 
be formed into a rectangle that the width 
will equal the line O, O. Process: First 
quadrature the rectangle ; you will then 
find by reference to Fig. J, the process 
for quadraturing a rectangle. 

By referring to Fig. H, you will find the 
rule for reducing a square to any rectangle 
you choose. 

When the rectangle is brought to its 
proper shape, the length will be equal to 
the distance from C to I, or set off the 
length of rectangle from C, on the side of 
triangle A, C, from C, and draw the line 

1, O, and you have your triangle halved. 



Problem No. 4. 

FIGURE J. 

Fig. J shows the method of changing a 
rectangle into a square. This rectangle 
is taken from Fig. I. One side of small 
triangle is taken for one side of rectangle. 
The other is the height of small triangle, 
this forms a rectangle of the same area as 
large triangle. Any triangle formed in 
this rectangle of the largest size that can 
be made in this rectangle, will be equal to 
the area of small triangle. To form it into 
a square, set your dividers at the angle 
B, open to C; this is the width of rectangle. 
Swing round to the side of rectangle, mak- 
ing the intersection at H This intersection 
will be one corner of square. Lengthen out 
the opposite side of rectangle at N. Place 
the corner of your square at this point H, 
in such a position as to make the inter- 
section at the end and side of rectangle of 
equal distance from the point H. This 
will give you two sides of square that will 
equal in area the rectangle. Proof: Form 
the square H, X, Y, N. This square 
takes in the rectangle with the exception 



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12 

of three triangles, but it takes in outside 
of rectangle three triangles of the same 
area and same shape; consequently the 
rectangle must be quadratured. This 
square being of the same area as rectangle, 
it follows that any triangle formed in this 
square that shall occupy one side of said 
square, and touch the opposite side, will 
contain the same area as triangle in rec- 
tangle. 



FIG.H 




Problem No. 5. 

FIGURE H. 

Fig. H shows the method of changing 
a square into a rectangle; the width being 
given, the square is of the same dimen- 
sions as square in Fig. J. For the width 
of rectangle we will take the length of 
line O, O, in Fig. I. Process: Draw two 
parallel lines apart the width required, 
set your dividers to the size of one side 
of square, then set one point in one 
of the parallel lines, and the other 
point in the other parallel line. Let 
these two points be two of the corners 
of square, and a line drawn from one 
to the other forms one side of square 
then form the square at the corner of 
square that comes between the two par- 
allel lines. Draw the line B, C, at right- 
angles with the parallel lines ; this gives 
you one end of rectangle required. Draw 
the line P, S, at right angles with parallel 
lines, and intersecting the corner of square 
at P. This gives you two triang'es out- 
side of the parallel lines and inside of 
square. Take the distance on parallel line 



14 



from the angle S to N, place it on the 
other parallel line from N. This will give 
you the other end oi rectangle, brect 
the perpendicular line D, O, and you 
have your rectangle, the area of which is 
equal to the square. Proof: The rec- 
tangle occupies the square with the ex- 
ception of three triangles, but it takes in 
three triangles of equal area. The length 
of this rectangle will be the distance re- 
quired from angle C, in figure I. 1 o 
find the angle I ot the halving line, draw 
the line from the point I to the given 
point in one side of triangle, and you 
have the triangle halved. Triangle D K, 
C, is the same as triangle in Fig. I, that 
halves the triangle from a given point. 



Problem No. 6. 

FIGURE G. 

Fig. G shows the method of halving 
triangles of a different shape than an 
equilateral triangle, and from either side 
of triangle, and by drawing the halving 
line parallel with either side of triangle. 
Let A, B, C be. a given triangle. To halve 
it by drawing the halving line parallel with 
the side A, B, first enclose the triangle 
with the rectangle A, B, E, C, as in figure 
G. Half of the square of the height of 
triangle taken from C, will be the halving- 
line. Process : Set one point of your 
dividers at the angle E, open to theangle 
C, swing round intersecting side of rec- 
tangle at D ; from the point of intersec- 
tion draw the line C, D, and from the 
line C, I), draw the line H, E,and at right 
angles with C, D, and intersecting at angle 
E. The length of this line is half of the 
square of the height of triangle. From 
the height of triangle and from C, take 
the length of this line, and you have the 
place for the halving line. Now draw the 
line O, O, parallel to the side of triangle 



i6 

A, B, and you have the triangle A, B, O 
halved. Now draw the line N, O, at right 
angles with rectangle and to intersect the 
angle at O, and you have all of the tri- 
angles that touch the point or angle at 
C halved. Also the rectangle is halved. 
In this figure you have also four different 
shapes of quadrilaterals, and five different- 
ly shaped triangles. Of triangles we have 
A, B, C; D, C, B; D, C, A; A, C, E, and 
D, C, E ; also the large rightangled tri- 
angle C, A, B, all halved by halving the 
rectangle. The rectangle could have been 
halved as well by taking half of the square 
of the length of rectangle from the line 
C, N, A, and from C. The line from C. 
to N, is half of the square of the length of 
rectangle. Of quadrilaterals we have 
one rectangle and four rhomboids of 
different shapes. 





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Problem No. 7. 

FIGURE A. 

Fig. A shows one method of quadratur- 
ing a circle. 

First draw the circle, then draw the 
lines O, S, and D, C, at right angles 
through the center of circle. These lines 
we will call the four radiuses of the circle. 
Then set one point of your dividers in 
angle formed by radius crossing circle at 
R, open to the angle Q of same radius, 
swing round intersecting extended ar- 
dius S, O; make the point S. Then draw 
the tangent A, O, B, parallel to extended 
radiuses D, C. Then draw the lines S, 
B, and S, A, from point S through the 
angle of circle and radiuses at Q and R, 
extending these lines to tangent A, O, B. 
This forms an equilateral triangle, either 
side of which is equal to half of the cir- 
cumference of circle. Then draw the 
lines B 7 C, and A, D at right angles with 
tangent and intersecting the two angles 
of triangle and extending to the line of 
the two extended radiuses. 

This forms a rectangle of the same area 



i8 

as circle. Now you may square root this 
rectangle by taking half the length of the 
rectangle for one side of a new rectangle 
and twice the width for the other side. 
This operation forms a new rectangle of 
the same area as the first. Then proceed 
to square root it by following the direc- 
tions given in Fig. J. 

For square rootingarectangle: You will 
observe the small rectangle at the top of 
Fig. A, that 1 have square rooted. This 
little rectangle you will see is one- 
fourth of the size of large rectangle; con- 
sequently it must be the area of a quad- 
rant of the circle. This little square 
occupies the little rectangle with the ex- 
ception of three triangles, and takes in 
three triangles of the same area. Now 
this circle can be quadratured very easily 
and quickly by drawing the line E, U, at 
right angles with radius, and intersecting 
the angle of radius and circle at E, and 
extending to the side of triangle, making 
the angle U. Then set one point of your 
dividers in the point S, open to the angle 
U, swing round to V, intersecting radius. 
Where this curve line intersects the ra- 
dius is the side of a square. When 
formed, as you see, with the sides of equal 



19 

distances from the center X, you have 
the circle quadratured. See how this little 
square at the top of this figure compares 
with the one-fourth of this large square. 



Problem No. 8. 

FIGURE B. 

Fig. B shows the method of changing a 
given square into a circle of the same 
area. Let A, B, C, D, be a given square. 
Now draw the two lines F, E, X, O and 
W, X, D at right angles through the center 
of square. We will call these lines the 
four radiuses. Now set one point of your 
dividers in the angle O, open to E, swing 
round, intersecting the extended radius 
at P. From this intersection draw the two 
lines P, O, and P, E. This forms an equi- 
lateral triangle. At the point P, of the 
triangle set one point of your^dividers, 
open to the angle X, swing round 
intersecting the side of triangle at U ; 
then draw the line V, U, at right angles 
with the radius, and intersecting the cur- 
ved line at the side of triangle. Set one 
point of your dividers in the center of 
square, open to angle V, and you have 
the radius of a circle whose area is the 
same as the given square. Draw the cir- 
cle. 

Proof: Open your dividers the diame- 



21 

ter of circle, then set one point of your 
dividers in the angle of circle and radius 
at R, swing round intersecting the ex- 
tended radius at D; from this intersection 
draw the line D, G, intersecting the angle 
of circle and radius at R, and extending 
up to G ; then draw the tangent G, W, 
and at right angles with radius W,X. From 
the intersection of tangent with the line 
G, D, to the radius W, X, is one-fourth 
of the circumference of circle. Draw the 
line G, F, at right angles with tangent to 
intersect the extended radius at F. This 
forms a rectangle F, G, W, X, of the area 
of half of the circle. Divide the tangent 
in two equal parts, then draw the little 
rectangle at the top of this figure to equal 
one-half of rectangle F, G, W, X, and 
square root the rectangle, and you have a 
square that equals one-fourth of large 
square or a quadrant of the circle. 

I have square rooted this little rectan- 
gle, as you will see, which is equal to one- 
fourth of given square. 

Rule for getting the area of a circle: 
Multiply half of the circumference by the 
radius. 

Rule for finding the circumference of a 



22 

circle : Multiply the diameter by three 
and a seventh. 

Rule for finding half of the circumfer- 
ence of a circle: Multiply the radius by 
three and a seventh. 

Rule for finding the solid contents of a 
pyramid : Multiply the square of the 
base by one-third of the altitude. 



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Problem No. 9. 

FIGURE C. 

Fig. C shows the method of drawing 
three circles in a given triangle of un- 
equal sides. The triangle being given 
draw the three perpendicular lines from 
the three corners of triangle to the cen- 
ter of triangle. Then from this center and 
at right angles with these three lines draw 
a line to intersect the three sides of tri- 
angle at N, N, N, and from these angles 
at N, N, N, draw perpendicular lines to 
intersect the lines squared from at O, O, 
O. Connect these three angles by a line 
drawn from one to the other. This forms 
another triangle, to be centered the same 
as first triangle. From this center and at 
right angles with the sides of small tri- 
angle draw the three lines to intersect the 
three sides of first triangle at X,X,X. From 
this angle at X, X, X, draw perpendicular 
lines to intersect the three perpendiculars 
drawn from the three corners of first tri- 
angle. Where these perpendicular lines in- 
tersect each other is the center of cir- 
cles ; when drawn they will touch each 
other, and each circle will touch two sides 
of the triangle. 



FIG.D 




Problem No. io. 

FIGURE D. 

Fig. D shows the method of halving the 
area of this shaped figure by drawing a 
line through it parallel with the ends, the 
figure being given. Divide the length in 
four equal parts, by lines drawn parallel 
with the ends of figure, then draw the di- 
agonal lines F, D, L, E, twice the distance 
from the crossing of these diagonal 
lines to B. Set off from the angle H, to 
K. This will be the dividing line; or 
twice the distance from the angle A, to 
the angle C. Set off from the end I, will 
be the dividing line, or twice the distance 
from the angle A to angle O. Set off 
from O, will be the dividing line. 

Proof : Form the two ends into rec- 
tangles, then square root them and you 
will find them equal. 



Problem No. i i. 

Method for quadraturing a circle by 
figures : We will take a circle that is eight 
inches in diameter; first multiply eight by 
three and one-seventh, and we find that 

g we have 251. This being the cir- 

-« cumference, we will multiply it by 
— - half of the radius, which is two. This 

^ produces fifty and two-sevenths. This 
— * is to be square rooted. In order to 
2 5: do this each square inch must repre- 
sent 7x7=49, and multiply forty-nine by 

. fifty. Now we have f by | to be 

- Q added. This makes the number to 

be square rooted, in order to quad- 

^"5 rature this circle, 
-g )246 4 (49.6 3 8 

8 9 )~864 
801 
986) 6300 
59i6_ 
9923) 38400 
29769 
99268) 863100 
794144 
68956 



26 

After rooting the number 2464, we 
find the number to be 49.638. This must 
be divided by the same number that we 
multiplied the square inches by. 



7091 


7)49638' 


7091 


49 


7091 


63 


63819 


63 


49637 


8 


50,282,281 


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1 



After multiplying the root by itself we 
find that we have 50.282 thousandths. 
One thousand divided by seven, and quo- 
tient multiplied by two, is 2851. By this 

7)1000(142 P r( ? cess w ^ £ nd a j° ss 0{ c 
' 2 only 3f of thousandths of 

— -pr- a square inch. 
30 284 

28 if 

20 285+ 

6x2=12 

This shows that when you square root 
a number for the area, you must have two 
dimensions instead of one. 

This rule must be observed and fol- 



27 

lowed when you wish to root a number 
for the area of a circle. 

The rule that I have adopted in this 
example for quadraturing a circle has 
never been published or taught For that 
reason mathematicians have never been 
able to quadrature a circle. Now we will 
square root the area of this 8-inch circle 
by geometry, the area being 50$ : First, 
7x7^49, Now form the square 7 by 7 
inches. This leaves one square inch and 
one piece, f by f Now form this f by 1 
into a rectangle ^ by £ inch. Now pro- 
ceed to square root this rectangle by the 
rules in Problem No. 4 for square rooting 
a rectangle. This being formed into a 
square, place the square inch by the side 
of it and in such a position as to have the 
side of each square on a line. This line 
we will call the base line, from the angle 
formed by the two squares being placed 
side by side. To the opposite side of the 
square inch, and at the corner formed by 
the base line will be the side of a square 
of the same area as the two squares. 
Place this square by the side of the 7 by 
7 square, and on a line with one side, and 
from the angle formed on the side of 
large square to the opposite side of large 



28 

square, and at the corner formed by the 
base line will be the side of a square of 
the same area as the 8-inch circle, with- 
out any loss or gain. 



Problem No. 12. 

To get the area of a sphere or a 
globe: Get the circumference of the 
sphere that you wish to get the area of. 
Then multiply the circumference by the 
diameter and one-seventh, and you have 
the area of sphere. 



OPERATION. 




Diameter, 


7 




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The solidity of a 


22 


sphere is equal to its 


n 


surface multiplied by 


154 


a third of its radius. 


3+ 




6)1571 Surface, 




26^ 



Solidity, 183J 



GEOMETRICAL 



Square Root; 



A CIRCLE QUADRATURED. 



OTHER PROBLEMS. 



BY JSTOttJMU^JST CIR^IBIB. 



Chicago- 

1S7 South Green Street. 

i<?79. g 



